| Date 06/09/2010 |
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| Script S4_2_10.m | |||
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%=============================================
%Fabry-Perot 3 %============================================= % %the two lines of sodium (in microns) lambda1=0.58900; lambda2=0.58958; %the thickness t (in microns) t=2.0e+003; %the angles of incidence (in degrees and in radians) tetag=eps:0.001:1.35; teta=tetag*(pi/180); %reflexivity, refractivity and their squares ro=0.95; tau=1-ro; roq=ro^2; tauq=tau^2; %phase shifts for the two wavelengths k1=2*pi/(lambda1); fi1=(k1*2*t)*cos(teta); % k2=2*pi/(lambda2); fi2=(k2*2*t)*cos(teta); % %========================================================================= %orders m1MAX, m1MIN, m2MAX and m2MIN varying teta %for lambda1 m1MAX=k1*t*cos(teta)/pi; M1=k1*2*t*cos(teta)/pi; m1MIN=(M1-1)/2; %for lambda2 m2MAX=k2*t*cos(teta)/pi; M2=k2*2*t*cos(teta)/pi; m2MIN=(M2-1)/2; %the first two minima (when teta=0) m1primo=m1MIN(1) m2primo=m2MIN(1) % %========================================================================= %RELATIVE INTENSITY Ir1 and Ir2 % %preliminary calculi used in both cases num3=tauq; dena=1+roq; % %========================================================================= %RELATIVE INTENSITY Ir1 %========================================================================= % denb1=2*(ro.*cos(fi1)); denb11=dena-denb1; Ir1=num3./denb11; %======================================================= %Only the first 1001 elements of the array Ir1 are used %to find values and positions of maximum and minimum Ir1p=Ir1(1:1001); [Ir1max,p1max]=max(Ir1p) [Ir1min,p1min]=min(Ir1p) %the corresponding angles ang1max=tetag(p1max) ang1min=tetag(p1min) ord1max=m1MAX(p1max) ord1min=m1MIN(p1min) % %a countercheck: remember the conditions %of maxima and minima for fi %(see pag.174: the last formulae of Sec. 4.2.8) % %for Ir1max fi1max=ord1max*2*pi; Ir1maxpr=num3/(dena-2*ro*cos(fi1max)) %for Ir1min fi1min=(2*ord1min+1)*pi; Ir1minpr=num3/(dena-2*ro*cos(fi1min)) % %======================================================================= %RELATIVE INTENSITY Ir2 %======================================================================= % denb2=2*(ro.*cos(fi2)); denb22=dena-denb2; Ir2=num3./denb22; %======================================================= %Only the first 1101 elements of the array Ir2 are used %to find values and positions of maximum and minimum Ir2p=Ir2(1:1101); [Ir2max,p2max]=max(Ir2p) [Ir2min,p2min]=min(Ir2p) %the corresponding angles ang2max=tetag(p2max) ang2min=tetag(p2min) ord2max=m2MAX(p2max) ord2min=m2MIN(p2min) % %a countercheck: remember the conditions %of maxima and minima for fi %(see pag.174: the last formulae of Sec. 4.2.8) % %for Ir2max fi2max=ord2max*2*pi; Ir2maxpr=num3/(dena-2*ro*cos(fi2max)) %for Ir2min fi2min=(2*ord2min+1)*pi; Ir2minpr=num3/(dena-2*ro*cos(fi2min)) %========================================================================= % %========================================================================= %we check the fulfillment of the Rayleigh'criterion %angles difference between the first maximum and the next minimum Ray1=abs(ang1max-ang1min) %angles difference between the first maximum and the second maximum Ray2=abs(ang1max-ang2max) %Ray1 (0.399°) isn't equal to or less than Ray2 (0.282°) %========================================================================= % %plot of Ir1 and Ir2 plot(tetag,Ir1,'r-',tetag,Ir2,'b-') %the command axis for Fig.4.30 axis([0.2 0.8 0 1]) %the command axis for Fig.4.31 % axis([0.35 1 0 0.05]) title('Ir1(red) and Ir2(blue) varying tetag from 0.2° to 1° with t=2mm and ro=0.95') %============================================= % |
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